從根節點往下分別查找左子樹和右子樹的最大節點，再比較左子樹，右子樹，根節點的大小得到結果，在得到左子樹和右子樹最大節點的過程相似，因此可以采用遞歸的

//樹節點結構??

public?class?TreeNode?{??

????TreeNode?left;??

????TreeNode?right;??

????int?val;??

????TreeNode(int?val){??

????????this.val?=?val;??

????}??

}??

public?class?Solution?{??

????/**?

?????*?@param?root?傳入根節點?

?????*?@return?返回最大結果?

?????*/??

????public?static?TreeNode?maxNode(TreeNode?root){??

????????//當再無子節點，返回當前節點??

????????if?(root?==?null)?{??

????????????return?root;??

????????}???

??????????

????????TreeNode?left?=?maxNode(root.left);//遞歸得到左子樹最大值??

????????TreeNode?right?=?maxNode(root.right);//遞歸得到右子樹的最大值??

????????return?max(root,max(left,right));//返回左節點，根節點,右節點的最大值??

??????????

????}??

????public?static?TreeNode?max(TreeNode?a,TreeNode?b)?{??

????????if?(a?==?null)?{??

????????????return?b;??

????????}??

????????if?(b?==?null)?{??

????????????return?a;??

????????}??

????????if?(a.val?>?b.val)?{??

????????????return?a;??

????????}??

????????return?b;??

????}??

????public?static?void?main(String[]?args)?{??

????????TreeNode?t1?=?new?TreeNode(1);??

????????TreeNode?t2?=?new?TreeNode(-5);??

????????TreeNode?t3?=?new?TreeNode(3);??

????????TreeNode?t4?=?new?TreeNode(1);??

????????TreeNode?t5?=?new?TreeNode(2);??

????????TreeNode?t6?=?new?TreeNode(-4);??

????????TreeNode?t7?=?new?TreeNode(-5);??

????????t1.left?=?t2;t1.right?=?t5;??

????????t2.left?=?t3;t2.right?=?t4;??

????????t5.left?=?t6;t5.right?=?t7;??

????????System.out.println(Solution.maxNode(t1).val);??

????}??

}??

轉載于:https://www.cnblogs.com/Darkqueen/p/9165794.html

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